Given an array Arr of non-negative integers of size N. The task is to find the maximum possible xor between two numbers present in the array.
Example:
Input: Arr = {25, 10, 2, 8, 5, 3}
Output: 28
The maximum result is 5 ^ 25 = 28
Input: Arr = {1, 2, 3, 4, 5, 6, 7}
Output: 7
The maximum result is 1 ^ 6 = 7
Naive Approach: A Simple Solution is to generate all pairs of the given array and compute XOR of the pairs. Finally, return maximum XOR value. This solution takes
O(N^{2})
time.
Below is the implementation of the above approach:
C++
// C++ implementation
#include <bits/stdc++.h>
using namespace std;
// Function to return the
// maximum xor
int max_xor(int arr[], int n)
{
int maxXor = 0;
// Calculating xor of each pair
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
maxXor = max(maxXor, arr[i] ^ arr[j]);
}
}
return maxXor;
}
// Driver Code
int main()
{
int arr[] = { 25, 10, 2, 8, 5, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << max_xor(arr, n) << endl;
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the
// maximum xor
static int max_xor(int arr[], int n)
{
int maxXor = 0;
// Calculating xor of each pair
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
maxXor = Math.max(maxXor, arr[i] ^ arr[j]);
}
}
return maxXor;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 25, 10, 2, 8, 5, 3 };
int n = arr.length;
// Function call
System.out.println(max_xor(arr, n));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation
# Function to return the
# maximum xor
def max_xor(arr, n):
maxXor = 0
# Calculating xor of each pair
for i in range(n):
for j in range(i + 1, n):
maxXor = max(maxXor,
arr[i] ^ arr[j])
return maxXor
# Driver Code
if __name__ == '__main__':
arr = [25, 10, 2, 8, 5, 3]
n = len(arr)
# Function call
print(max_xor(arr, n))
# This code is contributed by 29AjayKumar
C#
// C# implementation of the approach
using System;
class GFG {
// Function to return the
// maximum xor
static int max_xor(int[] arr, int n)
{
int maxXor = 0;
// Calculating xor of each pair
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
maxXor = Math.Max(maxXor, arr[i] ^ arr[j]);
}
}
return maxXor;
}
// Driver Code
public static void Main()
{
int[] arr = { 25, 10, 2, 8, 5, 3 };
int n = arr.Length;
// Function call
Console.WriteLine(max_xor(arr, n));
}
}
// This code is contributed by AnkitRai01
Time Complexity:
O(N^{2})
, where N is the size of the array
Efficient Approach: The approach is similar to this article where the task is to find maximum AND value pair.
So the idea is to change the problem statement from finding maximum xor of two numbers in an array to -> find two numbers in an array, such that xor of which equals to a number X. In this case, X will be the maximum number we want to achieve till i-th bit.
To find the largest value of an XOR operation, the value of xor should have every bit to be a set bit i.e 1. In a 32 bit number, the goal is to get the most 1 set starting left to right.
To evaluate each bit, there is a mask needed for that bit. A mask defines which bit should present in the answer and which bit is not. Here we will use mask to keep the prefix for every number ( means by taking the ans with the mask how many bits are remaining from the number ) in the input till the i-th bit then with the list of possible numbers in our set, after inserting the number we will evaluate if we can update the max for that bit position to be 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the
// maximum xor
int max_xor(int arr[], int n)
{
int maxx = 0, mask = 0;
for (int i = 30; i >= 0; i--) {
// se will store prefixes till ith bit for all the
// elements of array arr
unordered_set<int> se;
// set the i'th bit in mask
// like 100000, 110000, 111000..
mask |= (1 << i);
for (int i = 0; i < n; ++i) {
// Just keep the prefix till
// i'th bit neglecting all
// the bit's after i'th bit
se.insert(arr[i] & mask);
}
int newMaxx = maxx | (1 << i);
for (int prefix : se) {
// find two pair in set
// such that a^b = newMaxx
// which is the highest
// possible bit can be obtained
if (se.count(newMaxx ^ prefix)) {
maxx = newMaxx;
break;
}
}
}
return maxx;
}
// Driver Code
int main()
{
int arr[] = { 25, 10, 2, 8, 5, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << max_xor(arr, n) << endl;
return 0;
}
Java
// Java implementation of the above approach
import java.util.*;
class GFG {
// Function to return the
// maximum xor
static int max_xor(int arr[], int n)
{
int maxx = 0, mask = 0;
HashSet<Integer> se = new HashSet<Integer>();
for (int i = 30; i >= 0; i--)
{
// set the i'th bit in mask
// like 100000, 110000, 111000..
mask |= (1 << i);
for (int j = 0; j < n; ++j)
{
// Just keep the prefix till
// i'th bit neglecting all
// the bit's after i'th bit
se.add(arr[j] & mask);
}
int newMaxx = maxx | (1 << i);
for (int prefix : se) {
// find two pair in set
// such that a^b = newMaxx
// which is the highest
// possible bit can be obtained
if (se.contains(newMaxx ^ prefix))
{
maxx = newMaxx;
break;
}
}
// clear the set for next
// iteration
se.clear();
}
return maxx;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 25, 10, 2, 8, 5, 3 };
int n = arr.length;
// Function call
System.out.println(max_xor(arr, n));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the above approach
# Function to return the
# maximum xor
def max_xor(arr, n):
maxx = 0
mask = 0
se = set()
for i in range(30, -1, -1):
# set the i'th bit in mask
# like 100000, 110000, 111000..
mask |= (1 << i)
newMaxx = maxx | (1 << i)
for i in range(n):
# Just keep the prefix till
# i'th bit neglecting all
# the bit's after i'th bit
se.add(arr[i] & mask)
for prefix in se:
# find two pair in set
# such that a^b = newMaxx
# which is the highest
# possible bit can be obtained
if (newMaxx ^ prefix) in se:
maxx = newMaxx
break
# clear the set for next
# iteration
se.clear()
return maxx
# Driver Code
arr = [25, 10, 2, 8, 5, 3]
n = len(arr)
# Function call
print(max_xor(arr, n))
# This code is contributed by ANKITKUMAR34
C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to return the
// maximum xor
static int max_xor(int[] arr, int n)
{
int maxx = 0, mask = 0;
HashSet<int> se = new HashSet<int>();
for (int i = 30; i >= 0; i--) {
// set the i'th bit in mask
// like 100000, 110000, 111000..
mask |= (1 << i);
for (int j = 0; j < n; ++j) {
// Just keep the prefix till
// i'th bit neglecting all
// the bit's after i'th bit
se.Add(arr[j] & mask);
}
int newMaxx = maxx | (1 << i);
foreach(int prefix in se)
{
// find two pair in set
// such that a^b = newMaxx
// which is the highest
// possible bit can be obtained
if (se.Contains(newMaxx ^ prefix)) {
maxx = newMaxx;
break;
}
}
// clear the set for next
// iteration
se.Clear();
}
return maxx;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 25, 10, 2, 8, 5, 3 };
int n = arr.Length;
// Function call
Console.WriteLine(max_xor(arr, n));
}
}
// This code is contributed by Princi Singh
